Question

You are standing above the point (2,4) on the surface z=15−(3x

2
+2y
2
). (a) In which direction should you walk to descend fastest? (Give your answer as a unit 2-vector.) direction = (b) If you start to move in this direction, what is the slope of your path? slope = The temperature at any point in the plane is given by T(x,y)=
x
2
+y
2
+3
100
​
. (c) Find the direction of the greatest increase in temperature at the point (−2,2). What is the value of this maximum rate of change, that is, the maximum value of the directional derivative at (−2,2)? (d) Find the direction of the greatest decrease in temperature at the point (−2,2). What is the value of this most negative rate of change, that is, the minimum value of the directional derivative at (−2,2)?

Answer 1

To descend fastest from points (2,4), move in the direction of (-0.6, -0.8) with a slope of 20. The direction of the greatest temperature increase at (-2,2) is (-0.577, 0.577) with a maximum rate of change of 0.0566, and the direction of the greatest decrease is (0.04, -0.04) with the same rate in the opposite direction.

Step-by-step explanation:

To find the direction in which to walk to descend fastest from point (2,4), we should move in the direction opposite to the gradient of the surface z=15-(3x² + 2y²). Calculating the gradient, ∇z = (-6x, -4y), at the point (2,4) gives us ∇z(2,4) = (-12, -16). The unit vector in the direction of the steepest descent at (2,4) is therefore in the direction of the normalized gradient, which is approximately (-0.6, -0.8). The slope of the path as you start to move in this direction is the magnitude of the gradient at that point, which is √(12² + 16²) = 20.

The direction of the greatest increase in temperature at the point (-2,2) is given by the gradient of the temperature function T(x,y) = (x² + y² + 3)/100. The gradient here is (x/50, y/50), so at (-2,2), it is (-0.04, 0.04). The unit vector in the direction of the greatest increase is approximately (-0.577, 0.577). The maximum value of the directional derivative (the maximum rate of change) is the magnitude of the gradient at (-2,2), which is √((-0.04)² + (0.04)²) = 0.0566.

The direction of the greatest decrease in temperature at the point (-2,2) is the opposite direction of the gradient, which gives us the vector (0.04, -0.04). The minimum value of the directional derivative (the most negative rate of change) is the same as the maximum value but in the opposite direction, -0.0566.